Problems with Maxima and Minima

         Influenced by the continuous result of a square in exercises that ask for a maximum area, given a perimeter or minimum perimeter, given the area, I have determied a general equation for calculating these. The questions require that each side of a rectangle be represented in terms of one side given the length x.  This is solved as follows for a general perimeter (Formula 1) and area (Formula 2).

Formula 1

         Given the Perimeter of a rectangle P, find the Maximum Area enclosed by the rectangle.

                   Let the length and width of the rectangle be x and y respectively

                   P = 2x + 2y

                   y = P / 2 - x                                      ---(1)

                  A = x(P / 2 - x)

                 A' = P / 2 - 2x

                 Maximum value is at the stationary point (A' = 0)

                   x = P / 4                                            ---(2)

                   y = P / 2 - P / 4                (from (1) and (2))

                      = P / 4

                ∴ Maximum Area is given by a square with sides P / 4  (A = P² / 16)

         Note: the stationary point could also be found using the axes of symmetry, x = -b / 2a, of a parabola

Formula 2

          Given the Area of a rectangle A, find the Minimum Perimeter that encloses this area.

                     Let the length and width of the rectangle be x and y respectively.

                     A = xy

                      y = A / x

                     P = 2A / x + 2x

                    P' = -2A / x² + 2

                    Maximum value is at the stationary point (P' = 0)

                    2x² = 2A

                        x = √A

                        y = A / √A

                           = √A

                   ∴ Maximum Perimeter is given by a square with sides equal to √A (P = 4√A)